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June 2006 p4 q5
3895
Particles P and Q are attached to opposite ends of a light inextensible string. P is at rest on a rough horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. Q hangs vertically below the pulley (see diagram). The force exerted on the string by the pulley has magnitude \(4\sqrt{2}\) N. The coefficient of friction between P and the table is 0.8.
Show that the tension in the string is 4 N and state the mass of Q.
Given that P is on the point of slipping, find its mass.
A particle of mass 0.1 kg is now attached to Q and the system starts to move. Find the tension in the string while the particles are in motion.
Solution
(i) The tension \(T\) in the string is given by the horizontal component of the force: \(T = 4\sqrt{2} \cos 45^\circ = 4\) N. The mass of Q is found using \(T = mg\), so \(4 = 0.4g\), giving the mass of Q as 0.4 kg.
(ii) For P on the point of slipping, the frictional force \(F\) equals the tension \(T\). Using \(F = \mu R\) and \(R = m_P g\), we have \(4 = 0.8 \times m_P \times 10\), giving \(m_P = 0.5\) kg.
(iii) When a particle of mass 0.1 kg is attached to Q, the system starts to move. Applying Newton's second law to P and Q, we have: \(T - 0.8 \times 0.5g = 0.5a\) and \(0.5g - T = 0.5a\). Solving these equations gives the tension \(T = 4.5\) N.